Topic: Strength of Materials
A torque T is applied at the free end of a stepped rod of circular cross-sections as shown in the figure. The shear modulus of the material of the rod is G. The expression for diameter 'd' to produce an angular twist $\theta$ at the free end is
- A. $(\frac{30TL}{\pi\theta G})^{1/4}$
- B. $(\frac{33TL}{\pi\theta G})^{1/4}$
- C. $(\frac{18TL}{\pi\theta G})^{1/4}$
- D. $(\frac{24TL}{\pi\theta G})^{1/4}$
Answer: B. $(\frac{33TL}{\pi\theta G})^{1/4}$
Explanation: Total angle of twist $\theta = \theta_{1} + \theta_{2} = \frac{TL_{1}}{GJ_{1}} + \frac{TL_{2}}{GJ_{2}}$. For part 1: length $L$, diameter $d$, so $J_{1} = \frac{\pi d^{4}}{32}$. For part 2: length $L/2$, diameter $2d$, so $J_{2} = \frac{\pi (2d)^{4}}{32} = \frac{16\pi d^{4}}{32}$. $\theta = \frac{T \times L}{G \times \frac{\pi d^{4}}{32}} + \frac{T \times (L/2)}{G \times \frac{16\pi d^{4}}{32}} = \frac{32TL}{\pi G d^{4}} + \frac{16TL}{16\pi G d^{4}} = \frac{32TL}{\pi G d^{4}} + \frac{TL}{\pi G d^{4}} = \frac{33TL}{\pi G d^{4}}$. Solving for $d$: $d = \left(\frac{33TL}{\pi \theta G}\right)^{1/4}$.